Suppose that an object is at position s(t)=t^2 feet at time t seconds.

Now,I want to show you guys some physics and mathematics.

acceleration is the change in velocity in a given amount of time.\(a=\frac{V-Vo}{t}\)

where V is the final veocity and Vo is the initial velocity.

rearrange,we have \(V=at+Vo\)If the object's accelration is constant,then the graph of the velocity is a linear function.

But I would like to express it in y=mx+b form in my graph.And here is my graph

The area of the triangle is the sum of the  two base times the hight divide by two.or A=(b1+b2)*h/2

In here ,we could express it as \(A=(V+Vo)*\Delta t/2\)

the area under the curve(line) of function of velocity is the displacement.

Therefore,\(\Delta S=(V+Vo)*\Delta t/2\)\(a=\frac{V-Vo}{t}\Rightarrow t=\frac{V-Vo}{a}\)

If the object start moving at time 0 seconds 

\(\Delta t=T-To=t\)

the displacement of the object from time 0 seonds is \(\Delta S=\frac{V^2-Vo^2}{2a}\)

substitute V=at+Vo into \(\Delta S=(V+Vo)*\Delta t/2\)

we have \(\Delta S=(2Vo+at)*t/2\Rightarrow \Delta S=1/2at^2+Vo*t\)

displacement equal final position subract inital position \(\Delta S=S-So\)

\(S=\Delta S+So\)\(S=1/2at^2+Vo*t+So\)

Confirm my previous equation by using integration

\(V=\frac{ds}{dt}=a*t+Vo \Rightarrow \int \frac{ds}{dt}dt=\int at+Vo dt\Rightarrow S=1/2at^2+Vo*t+C\) for some costant C

In here C is the initial position,so \(S=1/2at^2+Vo*t+So\)

http://www.desmos.com/calculator/ibbxgibq5n (my original graph from demos)

The instantaneous velocity is  the derivative of the position function s=f(t) with respect to time.At tiem t,the velocity is 

V(t)=\(\frac{ds}{dt}=\lim_{x\rightarrow \Delta t } \frac{S(t+\Delta t)-S(t)}{\Delta t} \)

(1)v(t)=\(\frac{ds}{dt} t^2=2t\)

(2)\(\lim_{t\rightarrow 2} 2t=2*2=4\)

not good at calculus,might be wrong

Thanks fiora for your marvelously detailed answer :)

I think you should give yourself some points !

I have put your answer aside.  I would like to think about it some more.   ://